If $Y\sim\operatorname{Beta}(a,1-a)$ and $Z\sim\operatorname{Exp}(1)$, then $YZ\sim\operatorname{Gamma}(0,1)$?

Use the following result:

Assuming $Y$ and $Z$ are independent, the PDF of $X = YZ$ is given by:

$$f_X(x) = \int_{-\infty}^{\infty} \frac{1}{|u|} f_{Y}(u) f_Z\left(\frac{x}{u}\right) du$$

Here is a very familiar approach; nothing special about it.

Joint pdf of $(Y,Z)$ is $$f_{Y,Z}(y,z)=\frac{e^{-z}y^{a-1}(1-y)^{-a}}{\Gamma(a)\Gamma(1-a)}\mathbf1_{0<y<1,z>0}\quad,\,0<a<1$$

You can use a change of variables $(Y,Z)\to (U,V)$ such that $U=YZ$ and $V=Z$.

So the preimages are $z=v$ and $y=u/v$, and $0<y<1,z>0\implies 0<u<v$.

Absolute value of jacobian of transformation is $1/v$.

This gives the joint pdf of $(U,V)$:

$$f_{U,V}(u,v)=\frac{e^{-v}u^{a-1}(v-u)^{-a}}{\Gamma(a)\Gamma(1-a)}\mathbf1_{0<u<v}$$

Therefore, marginal pdf of $U$ is $$f_U(u)=\frac{u^{a-1}}{\Gamma(a)\Gamma(1-a)}\int_u^\infty e^{-v}(v-u)^{-a}\,dv\,\mathbf1_{u>0}$$

Substitute $v-u=t$, which converts the integral to a Gamma function, ultimately giving the answer $$f_U(u)=\frac{1}{\Gamma(a)}e^{-u}u^{a-1}\mathbf1_{u>0}$$